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I can visualize the exponential function with the graph $y = e^x$, but I can do that for almost any function.

In addition to its graph, the function $f(x) = x^n$ can be visualized as the volume of a box with sides of length $x$ in n-dimensional space, and the trigonometric functions can be interpreted as side lengths of certain right triangles.

Is there a similar geometric interpretation of the exponential function?

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if you want an $n$-cube of volume $(1+x/n)$ its sides are about length $e^x$...if you want to throw $\sin x, \cos x,\sinh x,\cosh x,e^x$ all in one bucket you can look at solutions to simple 1st/2nd order linear diff eqns...$x'-x=0,x''\pm x=0$ – yoyo Mar 25 '11 at 18:56
Further to your example of $x^n$, you could consider $\mathrm{e}^x$ as interpolating the volume of a box with sides of length $\mathrm{e}$ in $x$-dimensional space for non-integer dimensions $x$ -- but something tells me that might not be the sort of thing you're looking for :-) – joriki Mar 25 '11 at 19:36
This is not an answer to this question, but maybe it's useful nevertheless: the exponential function is characterized as the only continuous group homomorphism from $(\mathbb{R}, +)$ onto $(\mathbb{R}^+, *)$ such that its derivative at the origin is $1$. This is (IMHO) the best way to define and visualize it. – Giuseppe Negro Mar 25 '11 at 19:39
up vote 4 down vote accepted

There's a geometric interpretation of the natural log. From the definition

$$ \log x = \int_1^x {1 \over t} \: dt $$

we see that the area between the "standard" hyperbola $xy = 1$ and the horizontal axis between $1$ and $x$ is $\log x$.

So, turning this around, the line $x = e^t$ is the vertical line such that the area between $x = 1$ and $x = e^t$, between this hyperbola and the $x$-axis, is $t$.

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This answer is not about $e^x$, but about the closely connected hyperbolic functions $$ \cosh x=\frac{e^x+e^{-x}}{2},\quad\sinh x=\frac{e^x-e^{-x}}{2}. $$ A little algebra shows that $\cosh^2x-\sinh^2x=1$. Thus, $(\cosh x,\sinh x)$ are points in the hyperbola $x^2-y^2=1$; then $\cosh x$ and $\sinh x$ are the legs of a right triangle whose hypotenuse is the segment joining the origin and the point with coordinates $(\cosh x,\sinh x)$.

A different interpretation is the following: Imagine that you are running away fron some fixed point $O$. If your speed at each moment is equal to the distance to the point $O$, then your speed will be $C\,e^x$ for some constant $C>0$.

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If what you're looking for is a visual way to understand it, while a geometric interpretation might be useful, the most intuitive way to understand $e^x$ for me was stuff like population dynamics.

If you've ever heard the phrase "exponential growth" in relation to a population of something (bacteria, fish, wildlife, etc), it refers to the idea that each change in the population is linearly proportional to the population at the last time step -- assuming no outside influences in any way hinder or alter the population dynamics.

It stems from what I think is probably the simplest non-trivial ordinary differential equation there is:

$y'(t) = y(t)$

That is, there exists some function $y(t)$ such that $y(t)$ is the same function as its first derivative (up to a constant). The only function satisfying this criteria is $e^x$.

The implication may not be immediately obvious, so I hope you'll forgive the example.

If a population of fish is accurately described by $y$, then the larger my fish population is the faster the population will grow -- exponentially so. At each iteration, assuming a fixed average spawn-rate-per-fish, then the new fish added to the population will be a percentage of the size of the current population, and since the population grows every round then the increase and the population both grow at about the same pace.

This is true of financial investments as well that are strictly based on a fixed interest rate.

Of course, these are simplistic models. Fish populations don't really grow exponentially and financial investments don't really produce exponential growth.


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